text:
Why safe take off weight exceeds the safe landing weight?
This looks contra-intuitive: if a plane can take off and fly horizontally, what prevents it from descending and touching down to the runway with the required small vertical speed?

As long as there are human pilots, there will be times when the vertical speed when tire meets runway will not be small, whether from not flaring soon enough and flying into the runway or flaring too soon, holding it off too long, and then dropping it in. The typical landing is not a great landing. Indeed, there are times when you want to "plant it". Heavy crosswinds and short fields come to mind.

@Terry or (as I've seen happen) when misreading the runway elevation figure on the approach chart. Still see that 747-200C drop the last 10m or so after engagin TR in flight over the runway treshold when the pilots obviously noticed EHAM is at -12 feet rather than +12...

@jwenting I'm not familiar with the incident you speak of. If "TR" refers to thrust reversers, in theory it's not possible to to engage the thrust reversers while airborne as there's a lockout that prevents thrust reverser deployment until landing gear sensors report the aircraft is on the ground. Also, the flying pilot would never (or should never) even start into reverse until he feels contact with the runway.

@jwenting Ran out of characters, so continuing here. The last 100 feet or so is flown with reference to the radar altimeter, which of course reports AGL rather than MSL. It starts "talking" to you at 50 feet as I remember, reporting in 10 foot increments. My technique was to raise the nose 2 degrees above what I was holding when I heard the 50 foot call, then raise it another degree when I heard the 10 foot call. The 747 was the easiest airplane I ever flew in which to get consistently good landings. Of course, there are an unlimited number of ways to screw up.

@Terry I know the theory, I also know what I saw... Yes, he engaged reverse thrust while still in the air. May have been a glitch, an override of sorts, or deliberately disabled, but the effect was rather dramatic. Let's just say he made a very sudden final descent and landed with a mighty thud (but no damage severe enough to prevent him from taxiing to his parking slot, guess they did do a thorough check on the landing gear though).

what did you actually see when you say you saw the reversers engaged?

@terry Not just human pilots, but even automated systems will make hard landings occasionally due to weather conditions (wind gusts, thermals, etc.)

@Lnafziger I agree. Moreover, None of automated landings that I did in 747-100 and -200, which numbered probably only around a dozen or so since we only did them to see what would happen, were as smooth as any of us typically did by hand. The automated landings weren't hard landings, but I would describe them as clunkers.

Because when taking off you go from 0 vertical speed to a positive rate of climb all by way of the wings (the force on the gear will decrease over time to 0 when airborne). But while landing you will go from negative rate of climb to 0 vertical speed. Most of that will be absorbed by the gear (the force on the gear will spike and then settle to only the weight of the aircraft). After touching down there is also the braking force exerted on the gear which is proportional to the mass of the aircraft.
It is that spike+braking force that the engineers are worried about when calculating max landing weight.

I would add that on many aircraft overweight landing means the fire engine has to come around and cool the brakes as they will be red hot. 900°C not unheard of. This also happens when aborting take-off near V₁. The reason is both weight and the fact that at higher weight the speed is higher too.

The auto-ignition point of Kerosene (Jet Fuel) is 220°C.  So 900°C brakes could be very dangerous in the event of a leak or crack in the wing.  Of course, there would have to be another problem for it to actually pose a threat... but you get the point.

@SpongeBob: 900°C degree brakes are dangerous. 900°C is the highest number I ever saw mentioned, but the hot brakes warning illuminates at 300°C and overweight landings or high-speed rejects exceed that almost always and there are videos where the brakes visibly glow red which only occurs above 600°C or so.

The two weights are not directly related: in theory max take-off COULD be less than max landing, but that would show an issue with the design.
Take-off is determined by available field length in effect, and thus the engine thrust needed to get the plane airborne (before you run out of field). And in fact, you need a long enough field so that at V1 you can stop and and not fall off the end.
So the designer has to have enough thrust to goose it into the air fast enough, with enough fuel to fly to where they need to get to. More engines is more mass, more expense, and more fuel to run them.
Landing is not determined by engine thrust, in fact engine size works against you a bit because bigger engines are heavier. Landing relies only on drag and brakes. In fact you CAN land with max take-off weight, but that's an extreme, and you might burn your brakes/blow tires. Max landing weight also has to take into account bad weather, which increases required runway length. Also has to take into account possibility you might not have full use of flaps for drag, so need combination of longer runway, and enough braking capacity.
Commercial jets are all about economics and the trade-offs made in the design. You don't NEED to have max landing weight the same as max take-off, except in case of an emergency right after takeoff.
If you download PIANO (the free/trial version) you can select various weights for take-off and landing. See what it does to the landing field lengths.
Remember also that there is an existing set of airports that the designers want their planes to be able to take-off from, and land at. Also that the routes they want to fly have certain distances, needing enough fuel to get there.
As you increase take off weight, you need more lift, and more thrust. And you have to do it in the available runway length. When you land, you can take advantage of that lift (since you got it from a bigger effective wing area) but you also need brakes. Brakes are heavy and cost money to build as well as maintain.
Take off weight will vary from low (just enough to get to a nearby airport) to the max: enough to get to the farthest one. Landing weight SHOULD ideally be the same: just the empty plane weight and the reserve. Anything more and something has gone wrong. So the designer aims for that, and builds enough safety into the system to allow a landing with max takeoff, but with the possibility of tires blowing, or at least having to wait until brakes cool.
I read an account of commercial planes flying into Kabul fully loaded with fuel for re-supply, and they had problems with hot brakes, blown tires etc. Commercial planes are not designed for that. True military tankers/freighters have more effective reverse thrust, and need to use it to land on short runways. And there is no concern for passenger safety for hard deceleration, nor for noise.
The main reason commercial jets don't use reverse thrust more is because of noise (and emission) regulations. Otherwise you could land as heavy as you wanted: reverse thrust can give you up to 20%-30% of max thrust, and is effective at high speeds.
And if you look at engine thrust, there is a huge difference between take-off max thrust and cruising. Once you are airborne, you don't need a very big engine. But you have to carry around all the extra bulk just to get into the air.
On short flights, one of the big issues is the time to cool the brakes. If you have a quick turn-around and not much time between your next landing then your brakes don't have much time to cool. In that case the max plane weights are not really the issue: you are not carrying much fuel.
Note: Airbus has data on how much reverse thrust each plane has. It is a function of N1 (fan speed) and air speed. See:
Airbus Takeoff Safety

On the other hand, bigger engines also have a larger frontal area (and thus produce more drag), and have more potential reverse thrust available (although aircraft do have to be certified to land safely without reversers, in normal operations, most to all landings will use the reversers).

"True military tankers/freighters have more effective reverse thrust, and need to use it to land on short runways."  Short runways, or wet runways, or icy runways (even for commercial aircraft, a typical stopping distance on an icy runway with no thrust reversers is about three times as long as a typical full-reverser stopping distance)...

Ratchet freak's answer is entirely correct, but I would like to add to it. A plane could have a max landing weight that is equal or greater to the max take off weight. There is no technical limitation to it: the landing gear can simply be made to be sturdier.
However, sturdier landing gear is heavy and expensive. A plane will (usually) be lighter when it lands than when it takes off, because it burned fuel, so the additional sturdiness will seldom (if ever) be used. Therefore, manufacturers decide to make planes that have a safe take off weight that exceeds the safe landing weight.

I think another factor is that there are at least two ways "safe landing weight" could be defined, and the "safe take-off weight" is between them.  The upper definition for "safe landing weight" would be the maximum weight that would not significantly increase the risk of injury to passengers upon landing.  The lower definition would be the maximum weight that would leave the plane in presumably-flight-worthy condition without requiring a detailed inspection and/or overhaul.  Landing while overweight may not be "dangerous", but would necessitate extra inspections before the next take-off.


text:
What rules would allow a pilot to refuse a passenger's boarding?
I would like to know about any rules that would authorize a pilot to refuse to allow a specific passenger from boarding his plane (i.e. mental instability, obvious sickness, drunkeness, etc.).
Hypothetically, if someone were being 'forced' take a flight they did not want, what would be the options for the passenger to be refused boarding? I am only interested in the rules for a pilot.

The regulation that springs immediately to mind is FAR 91.17(b):

Except in an emergency, no pilot of a civil aircraft may allow a person who appears to be intoxicated or who demonstrates by manner or physical indications that the individual is under the influence of drugs (except a medical patient under proper care) to be carried in that aircraft.

i.e. "If you get yourself blind stinking drunk at the airport bar, or you've been doing lines of cocaine in the airport bathroom you are not getting on the plane because the FAA says I can't let you."

For other cases you need to head over to 49 USC 44902, which deals with Refusal to transport passengers and property for "air carriers" (FAR part 121, and I believe 135 operations are covered by this bit of the US Code). The most relevant text is:

(b) Permissive Refusal.
  Subject to regulations of the Under Secretary, an air carrier, intrastate air carrier, or foreign air carrier may refuse to transport a passenger or property the carrier decides is, or might be, inimical to safety.

Most air carriers vest the captain with the decision-making authority here: If the captain determines that a passenger may be "inimical to safety" they can refuse to fly the passenger.  There are some restrictions on that authority - notably that the basis for the decision can't be "discriminatory" - which are scattered about in 14CFR (the FARs) and 49USC (Transportation), as well as case law. The FAA has issued at least one legal interpretation discussing this in more detail.

For commercial (or non-commercial) operations under part 91 I wasn't able to find anything specific in 14CFR or 49USC, but in those cases the general consensus seems to be that the PIC has at least the level of authority granted under 49USC44902 - if not broader authority (i.e. if you're going on a personal flight you could refuse to take someone along because you don't like their shoes, and if someone hires you for a commercial flight you could refuse the job if they were acting sketchy).

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Where is the 'zone of reversed commands...'?
There seems to be some confusion about where the 'zone of reversed commands' really is, i.e. where a decrease of airspeed results in an increase of drag.
Some references place that zone to the left of the minimum of the power required curve, while others place it (correctly, in my opinion) to the left of the minimum of the thrust required curve... Those minima are separated by a 30% difference in airspeed, hence the distinction isn't academic...
Two examples of these contradictory references, one taken from the internet, and the other from a book on gyros ('Flugphysik der Tragschrauber'):

notice both references put it at before minimum sink speed, or highest L/D, or "best endurance". Here, it is not contradictory, and applies to all aircraft.  In a glider, if you trimmed back to a lower airspeed, your sink rate increases, burning more "fuel" (altitude).

@RobertDiGiovanni. No... You are very wrong... The best endurance speed is NOT the speed for best L/D...            The minimum of the drag curve is indeed the airspeed for highest L/D, but the minimum of the power required curve marks a different airspeed, the 'best endurance speed', that is about 30% less than the airspeed for best L/D.

You wrote minimum sink speed, or highest L/D, or "best endurance"    For propeller propulsion, that  is wrong. Minimum sink speed is indeed the best endurance speed, but it's not the highest L/D speed...

we were both looking at the wrong curve.  In level flight, "best endurance" is lowest fuel consumption, not maximum power available, nor even most efficient generation of thrust per gallon (optimal PROP AOA). No, it's just thrust, which is (perhaps not linearly, but always proportional) to RPM, which is proportional to fuel burn.  So the curve to look at in this case is prop RPM. A minimum will be found at minimum sink rate.  Think if it this way, glide first, find min sink speed, add any engine you want.  In ALL cases (In level flight) you will burn more fuel going faster or slower.

Vorderseite = front side,  Ruckseite = back side

This is commonly known as the Speed Stability, not to be confused with speed stability in the sense of static longitudinal stability. I think there's some common confusion with this phenomenon. The first part of this answer addresses the OP's question; the second part clarifies a common confusion.
1. Main Answer
In level flight (i.e. zero vertical rate), the longitudinal equation of motion can be succinctly written as:
$$m\dot{V}=T-D$$
$m$ is airplane mass, $V$ is airspeed, $T$ is thrust and $D$ is drag. Now if we express thrust and drag as first order approximation as a function of change in airspeed ($\Delta V$) from the trimmed condition, we have:
$$T=T_0+\frac{dT}{dV}\Delta V=T_0+T_V\Delta V$$ and $$D=D_0+\frac{dD}{dV}\Delta V=D_0+D_V\Delta V$$
At trim condition, we necessarily have $T_0=D_0$. So now we have a new equation of motion:
$$m\dot{\Delta V}=(T_V-D_V)\Delta V$$
This equation is a first order ordinary differential equation, and is stable if $T_V-D_V<0$ and unstable otherwise.

For a jet plane, thrust is fairly constant in flat rated conditions and $T_V$ is approximately zero. Thus, $D_V=0$ corresponds exactly to minimum drag or minimum thrust required (where $C_{D_0}=C_{D_i}$ for high aspect ratio, low Mach airplanes).
For a propeller plane, power is constant, but now the stability criterion is $-\frac{P}{V^2}-D_V<0$. This corresponds neither to minimum power required nor minimum thrust required.

2. Addendum
What does this result mean, exactly? If an airplane is trimmed in the speed unstable regime, will it decay toward stall if it experiences a speed perturbation with pilot hands-off, even if it's statically longitudinally stable?
Remember, the equation we began with only holds in level condition where the airplane is neither climbing nor descending. Thus, the pilot must be holding altitude with elevator while the speed is changing. The conclusion for speed instability is:

If the altitude deviates below the trim altitude, pulling up will decrease the airspeed, resulting in further energy deficiency. Speed will continuously decay in this manner.
If the altitude deviates above the trim altitude, pushing down will increase the airspeed, resulting in further energy excess. Speed will continuously accelerate until reaching a stable point on the other side of the polar.

Since the pilot is in the loop, it has nothing to do with the basic aircraft eigenmodes.
The same result can be obtained if we aim for constant flight path angle. In approach configuration, this is also called flight path stability, which I think is a better name than speed stability.

May the downvoter please add a comment on what is perceived to be inaccurate?

hands off recovery from a pitch perturbation is a hallmark of static stability, which is a created with aerodynamic tail downforce (pitches up when plane too fast) and forward set CG (pitches plane down when plane too slow).  While this esoteric "speed unstable" "zone of reversed command" lingo makes for interesting theoretical discussion, there is no way I would even mention the terminology to a student pilot.  Flying at 35 knots indicated with full throttle kind of got the point across for me.

now on to your (good) work.  Power supplied is not constant, it is controlled by the throttle.  Efficiencies aside (for now), more thrust requires more RPM, more RPM requires more fuel.  The curve we should be looking at (for endurance) is again minimum thrust.  Do we have a case where lower prop RPM produces more thrust?  No.  So maximum endurance will be at the lowest RPM setting possible.  Honestly, the force x distance definition pains me. Power as force over time (fuel burn per hour) makes more sense.

@RobertDiGiovanni I don't understand what you're trying to get across here...

Your work is interesting. 2. Minimum thrust is proportional to throttle setting.  Maximum power available curve doesn't seem to have anything to do with this.

Both are right in their own ways.
First the basics: Power is thrust times speed and is constant over speed for variable-pitch propeller-powered piston aircraft. Therefore, propeller thrust is proportional to speed inversed. The minimum power required coincides with the point of maximum excess power since it needs the lowest power setting for trimmed flight. The minimum thrust required is when the absolute drag is at its minimum. Since variable-pitch propeller thrust is inverse to speed, flight at minimum drag needs more power and happens at a higher speed than trimmed flight at the maximum excess power speed in propeller-powered piston aircraft.

Thrust (green), drag (red) and excess power (blue) over speed for a propeller aircraft. The broken line is for trimmed flight at the lowest power setting. Numbers for a light GA aircraft with 106 kW and 1300 kg mass.
Now to the instability: Normally, if you increase speed you need more power to trim the aircraft at this higher speed. Without pilot intervention, the aircraft will slow down to the trimmed speed. This is a stable process. Works in reverse, too. In the plot above in the stable range, you can see that if you move away from the trim point, drag will drop less than thrust, so any speed increase will need a higher power setting and the aircraft falls back to the trim point. This works for every power setting to the right of the maximum excess power speed.
Below that point this condition inverses: Now drag will increase more with a speed decrease than thrust will increase. To the left of the maximum excess power speed, a deviation from the initial speed will either speed the aircraft up or slow it down with no hope of ever returning to the initial state without pilot intervention.
Obviously, the maximum excess power speed (which is the minimum power required speed and the best endurance speed for propeller aircraft) is the boundary between stable and unstable behavior. For propeller aircraft.
With turbojets, thrust is roughly constant over most of the subsonic speed range and now the maximum excess power speed coincides with the minimum thrust required speed. Your book deals with turbojets while those web pages explain speed stability for variable-pitch propeller aircraft.
Note that fixed-pitch propellers show a linear increase of efficiency over speed below their optimum advance ratio, so here again the thrust is approximately constant over speed. But that holds only over the slow speed range when the aircraft flies slower that what the propeller advance ratio would like.

I disagree with your characterization in the third paragraph. Speed instability (imo a terrible misnomer) has nothing to do with speed stability in the sense of static longitudinal stability.

@Jimmy: Look at the equations of motion and solve them for different speeds. The eigenvalues will produce an unstable condition for speeds below that instability boundary. This is the same kind of instability as the roll instability. And then change cg location and see what that does to the eigenvalues. After having done this, would you still think that both have noting in common?

I have done just that and I can confirm this is not true. Drag only improves phugoid stability, and short-period cannot go unstable without static long stab going unstable.

@Jimmy: When the cg is shifted aft to an unstable position, the complex eigenvalue pair of the phygoid splits into two real eigenvalues, one of which crosses into the positive side, signaling instability. Just what the speed stability eigenvalue does when speed drops below the stability limit. If this did not happen, you did something wrong. Drag is not an independent variable in this. The short period mode will split into two real pairs later, when cg is shifted behind the maneuver neutrality point.

I think may be erroneous to link the concept of the "region of reversed command" to pitch stability.  See my alternative answer.

"Below that point this condition inverses: Now drag will increase more with a speed decrease than thrust will increase. To the left of the maximum excess power speed, a deviation from the initial speed will either speed the aircraft up or slow it down with no hope of ever returning to the initial state without pilot intervention." -- seems to me that this misses a large part of pitch stability dynamics-- the way that a decrease in airspeed decreases the lift coefficient which makes the flight path TEMPORARILY curve more downward which makes the gravity vector exert a larger thrust compnent

@quietflyer: This is not pitch or longitudinal stability. This is speed stability.

Long comment immediately above should have read "decreases the lift vector" not "decreases the lift coefficient".  For example the effect would still apply even in response to a sudden wind gust producing a  sudden airspeed change even if the direction of the gust was such that there was no change in lift coefficient at all, at least at the first instant that the gust began.  And I'm still not seeing why an aircraft would no longer have "speed stability" in the "region of reverse command", when all relevant factors are considered.

@PeterKampf Speed stability has nothing to do with stall or separation induced pitch changes. It can be easily shown with induced drag alone.

One definition of the "region of reversed command" is the part of the flight envelope where a small aftwards movement of the stick or yoke, causing a small decrease in airspeed, with no change in the position of the throttle or thrust lever, will eventually lead to a net increase in sink rate, not a net decrease in sink rate.
In the "region of reversed command", you must advance the throttle or thrust lever, not retard it, to maintain altitude (or to maintain a constant climb rate or sink rate) as you slowly move the control stick or yoke aft to decrease airspeed.
Note that this does NOT mean that in the "region of reversed command", if you want to maintain a constant altitude as you move the throttle or thrust lever forward, you are forced to move the stick or yoke aft and let the airspeed decrease.  You could instead put the stick forward to allow the aircraft to accelerate out of the "region of reversed command".
If for a given position of the throttle or thrust lever, your aircraft's engine is putting out constant horsepower regardless of airspeed, which is characteristic of a piston or turboprop engine, then the "region of reversed command" will be the part of the flight envelope where the airspeed is lower than the airspeed for minimum power-required.  I.e. the part of the power-required graph that lies to the left of the airspeed where the minimum power is required.
On the other hand, if your aircraft is putting out constant thrust regardless of airspeed, which is characteristic of a jet engine with no propeller, then the "region of reversed command" will be the part of the flight envelope where the airspeed is lower than the airspeed for minimum thrust-required.  I.e. the part of the power-required graph that lies to the left of the airspeed where the minimum thrust is required.
A more nuanced approach would consider the shape of the power-required curve and the power-delivered curve for any given position of the throttle or thrust lever.  On such a graph, the "region of reversed command" is any region where, as we decrease airspeed, we make the value of (power delivered minus power required) get less positive or more negative.  This is the part of the flight envelope where a decrease in airspeed will lead to a decrease in climb rate or an increase in sink rate, with no change in the position of the throttle or thrust lever.
This answer could be expanded to consider the effects of constant-speed propeller.
All the above content could also be restated and simplified to eliminate any reference to the position of the control stick or yoke, and focus solely on airspeed.  Also if we wished, we could eliminate any reference to the position of the throttle or thrust lever, and focus solely on thrust-required or power-required.  It's just a matter of definitions.  If by "region of reversed command", we simply mean that we need more thrust to maintain altitude while flying a little slower than while flying a little faster, than obviously the "region of reversed command" is the part of the thrust-required graph that lies to the left of the minimum thrust-required point.  Similarly, if by "region of reversed command", we simply mean that we need more power to maintain altitude while flying a little slower than while flying a little faster, than obviously the "region of reversed command" is the part of the power-required graph that lies to the left of the minimum power-required point.  It's just a matter of defining our terms.
We could also define the "region of reversed command" slightly differently-- as the part of the flight envelope where a small aftwards change in the position of stick or yoke, leading to a small decrease in airspeed, causes the glide angle (relative to the airmass) to get steeper rather than shallower.  In the power-off case, this region would include any airspeed lower than the best L/D airspeed.
With any of these various definitions, it would seem to be an error to suggest that being in the "region of reversed command" profoundly changes an aircraft's basic pitch stability dynamics and/or pitch control response dynamics and/or speed stability dynamics, unless we've introduced an autopilot (or human pilot) into the loop that is trying to use pitch control inputs to maintain altitude or maintain a set climb or descent rate or stay on a fixed glide slope, etc.
Another answer has inspired the following thoughts:
In the "region of normal command", we may control the aircraft in any of the following ways:
1) Increase airspeed by moving the control stick or yoke forward, and decrease airspeed by moving the control stick or yoke aft, with no change to the position of the thrust lever or power lever (throttle).  The climb rate or sink rate will not remain exactly constant.
2) Increase airspeed by moving the control stick or yoke forward, and decrease airspeed by moving the control stick or yoke aft, while simultaneously adjusting the thrust or power level (throttle) as needed to hold altitude or climb rate or sink rate constant.  (Move the lever forward to increase climb rate or decrease sink rate, and aft to decrease climb rate or increase sink rate.)
3) Increase sink rate (or decrease climb rate) by moving the control stick or yoke forward, and decrease sink rate (or increase climb rate) by moving the control stick or yoke aft, with no change to the position of the thrust or power lever (throttle).  The airspeed will not remain exactly constant.
4) Increase sink rate (or decrease climb rate) by moving the control stick or yoke forward, and decrease sink rate (or increase climb rate) by moving the control stick or yoke aft, while simultaneously adjusting the thrust or power level (throttle) as needed to hold airspeed constant.  (Move the thrust or power lever forward to increase airspeed, and aft to decrease airspeed.)
In the "region of reversed command", only methods 1 and 2 will work.  Methods 3 and 4 will not work.
Other methods of controlling the aircraft that will work in the "region of reversed command"--
5) Move the control stick or yoke aft to increase the sink rate, and move the control stick or yoke forward to decrease the sink rate, while leaving the thrust or power level in a fixed position.  The airspeed will not stay exactly constant.  Do not try this near the stall angle-of-attack!
6) Move the control stick or yoke aft to increase the sink rate, and move the control stick or yoke forward to decrease the sink rate, while simultaneously adjusting the thrust or power level (throttle) as needed to hold airspeed constant.  (Move the thrust or power lever aft to increase airspeed, and forward to decrease airspeed-- very counterintuitive-- just as with method 4, this method only works because it forces the pilot to modulate his pitch control inputs in a way that leads to the desired change in airspeed.)  Again, it is unwise to try this near the stall angle-of-attack.
However, methods 5 and 6 will ONLY work if the pilot waits a good while after any pitch input to see the ULTIMATE, not IMMEDIATE, result in the sink rate or climb rate before making an additional follow-up pitch input.  Thus they are really not very practical in most cases.
To help illustrate the difference between the immediate and ultimate results of a pitch input in the "region of reversed command", consider this-- it is perfectly possible to execute a landing flare in the "region of reversed command", keeping the stick or yoke moving aft to generate an abnormally low sink rate (essentially zero or nearly zero) for the power setting (which may be near zero, especially in a light plane or glider.)  Here the stick is or yoke is being moved AFT, not forward, to arrest the sink rate as the airspeed decreases.  However, this maneuver is not sustainable-- if the pilot keeps moving the stick or yoke aft, the plane will eventually stall, while if he stops the aft movement of the stick or yoke just short of the stall position, the plane will start sinking at a high rate.
In the "region of reverse command", for maneuvers other than the landing flare, it is usually best to keep things simple and use pitch inputs ONLY for airspeed control, not control of sink rate or climb rate.  In the "region of reverse command", it is most practical to accomplish changes in sink rate or climb rate by moving the thrust or power lever.

I've now reworked my answer considerably.

I still don't agree that a glider, or any other aircraft, is going to experience "speed instability" while flying on the backside of the sink rate polar, unless it is caused by the way an autothrottle and/or autopilot is connected into the control loop.

In additions to variations 1 through 6 listed above, the answer could also consider variations based on holding a constant pitch ATTITUDE while manipulating the throttles or power levels to control sink rate or stay on a glide path.

Well, this answer is not from me, but I found it, minutes ago, in Richard von Mises' 'Theory of Flight' Dover Books, ISBN 978-0-486-60541-8.
'The abscissa where the two power curves have parallel tangents...'

According to Bold Method, and importantly pertaining to your checkride, the region of "reversed command" is essentially slow flight technique, where pitch controls speed and power controls altitude, as compared with cruising flight, where one "levels off" with pitch and "throttles back" to cruising speed. This, crucially, is technique used for approach and landing, where a safe speed can be trimmed in.
This is not a definition to be remembered by rote, as it only serves to confuse common sense:  that speed control is critical for slow flight and you can't "count" on your engine to save you if you "screw it up" (or down). Glider pilots know this well. Managing speed with pitch, and altitude with power, sounds very "normal" to me, and yes, looking at it properly, works the same way at cruise.

Folks, for any trim condition (staticly stable), a plane will climb if power is added and sink when power is taken out (from it's flight path).  Set that in stone.  In the region of "reversed command" moving the elevator trim for a lower airspeed (higher AOA for same lift) requires MORE throttle due to increased induced drag.  In the "normal" region moving elevator trim for lower airspeed requires LESS throttle due to less parasitic drag.  One can see how the least amount of throttle is required at minimum sink rate speed, aka "best endurance speed", aka highest L/D AOA.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
What is this substance on these pipe fittings?

In this picture from a different question I noticed some kind of colored substance across many of the fittings. It is in several different colors and looks like it might be sticky. (The pink kinda looks like chewing gum. That's even worse than the duct tape question. ;) I have circled some of them in yellow.
Is it there to mark whether the fittings have been tampered with?

Typically, each mechanic will have their own color of torque seal (it comes in a whole rainbow of colors). In that way, not only can you tell if a fitting has been tampered with, but who it was who last worked on that system.

That is called "torque seal" or "torque putty".
It is used by mechanics to mark a part after it has been properly torqued, and thereafter to serve as an indication if loosening has taken place. The torque seal hardens after application and serves as a marking mechanism that will show—by cracking and/or flaking of the putty—if the part has loosened resulting in rotation.
The photo below shows torque seal as initially applied (left), and then with the seal broken following loosening an rotation of the nut (right).

Image courtesy of theleanwarehouse
Just to be clear, the torque seal does nothing to secure the part or otherwise serve any locking function. It serves only as an indication.
Dykem is one manufacturer of torque seal.
Here is a page with additional information about torque seal from a retailer.

And if you run out you can just use juicyfruit :)

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How much lift is created by a half-sized rotor?
I wonder if lift is directly proportional to the diameter of the rotor disc.
For example, if an engine spinning a 26 ft diameter rotor can lift 635 kg, will a 13 ft diameter rotor using the same engine lift 318 kg? The power absorbed by the rotor is the same in both cases.

welcome to aviation.se this is a homework question, we do not do homework, but we help you understand. in order to receive a good answer can you specify what have you tried? what do you not understand?

sir i want to know that if a full sized rotor is lifting lets say 635kg weight at given rpm and power at a given altitude at ft/min than how much half sized rotor will lift at same rpm and power at same altitude at ft/min will it be half the weight like 635/2=318kg ?

Your question is very hard to read because you have not used spacing and paragraphs.  It is also hard to answer because an R22 cannot lift 635kg at 14,000 feet.

For the hovering rotor, the stationary case, it can be safely assumed that the lift $L$ is a function of the input power $P$, the diameter $D$ of the rotor and the air density $\rho$.
Thus, $L = f(P,D,\rho)$
where $f$ is a function to be determined.
From dimensional analysis, the lift $L$ can be easily derived:
The variables are Lift $L$, dimensions $MLT^{–2}$; Power $P$, dimensions  $ML^2T^{–3}$; Rotor diameter $D$, dimensions $L$ and air density $\rho$, dimensions $ML^{–3}$
The variables form a non-dimensional product $k$
$k = L^a\cdot P^b\cdot D^c\cdot \rho^d$  where $a,b,c,d$ are numbers to be determined.
Let’s form now a parallel product $k^*$ with the dimensions:
$k^* = (MLT^{–2})^a  (ML^2T^{–3})^b  (L)^c (ML^{–3})^d$
Clearly, $k^* = M^0 L^0 T^0$... We now take the exponents for each dimension:
$a  + b + d = 0 \\
a + 2b + c – 3d = 0 \\
–2a – 3b = 0$
We make $a = 1$, since $L$ is the variable we’re going to solve for.
$b = –2/3 \\
d = –1/3 \\
c = –2/3$
Then,
$k = L^a\cdot P^b\cdot D^c\cdot \rho^d \rightarrow  k = L\cdot P^{–2/3}\cdot D^{–2/3}\cdot \rho^{–1/3}$
Solving for $L$
$L = k\cdot P^{2/3}\cdot D^{2/3}\cdot \rho^{1/3}$
where $k$ is a constant
Hence, for rotor diameters $D_1$ and $D_2$, and for the same power and air density, the corresponding lifts $L_1$ and $L_2$ are:
$L_1/L_2 = (D_1/D_2)^{2/3}$
For the case of $D_1 = 13 ft$ and $D_2 = 26 ft$, $L_1/L_2 = (13/26)^{2/3} = 0,63$
In other words, the smaller (13 ft) rotor gives you, for the same power and air density, just 63% of the lift attained with the larger (26 ft) rotor.
That's for the hover. For the climb, you'll need extra power. In order to move 635 kg vertically upwards at 1200 ft/min (6,09 m/s) you would need $635 \cdot 9,8 \cdot 6,09 m/s = 37,9 kW...$

Nice math... Though I am not convinced. OP says same power, which would mean a much higher RPM and all the extra losses associated with that. I guess it depends on your definition of $P$.

@Trevor From the Froude-Rankine 'actuator disk' approach, a similar formula (but with the constant...) can be derived, expressing the thrust of a propeller as a function of the power applied to the prop, its diameter and the density of the fluid...

Yes, but I think it worth mentioning, switching rotors wont actually get you "the same power applied to the rotor". What the appropriate derate factor /RPM is, is of course is indeterminate.

@Trevor  Of course, the 'actuator disk theory' is not an exact solution, but just a useful, approximate approach...

For the smaller rotor: higher RPM or higher blade pitch?

No, it won't be half.  It will be much less than that.
The area of lift is circular, so halving the diameter decreases the area of the circle to 25% of the original.  Area = πr2.
Besides the effective rotor area, there are also losses for the body of the aircraft being in the center, though the rotor doesn't generate lift near the center because of low blade speed and not having the blades go all the way to the center.  I would look for a helicopter design rule-of-thumb to find the answer.

"I would look for a helicopter design rule-of-thumb to find the answer.": It's what the OP is asking: How to calculate the lift vs the rotor diameter, all other parameter unchanged?

Exactly that is what i want to know

The loading per blade of the rotor is roughly the same for a wide variety of helicopters. If the number of blades is kept constant, the possible lift of the half-sized rotor will be a quarter of the original rotor. Of course, it will also absorb less power. However, in this case the engine is the same, so @xxavier's conclusion gives a better answer. Ideally, the smaller rotor would have deeper and more blades.

@PeterKämpf But that wasn't part of the problem statement ... which leaves the answers having to guess at the nature of "all else being equal." (Your points are well taken).  It might be useful to engage with the OP and get a tad more definition for the question.

@KorvinStarmast: Yes, you're right. That's why I started the sentence with  "Ideally". Anyway, this smells a lot like some homework question, so my level of engagement is low.

@PeterKämpf Read you loud and clear. :-)

The original problem did say "same rpm and power," which I think rules out any possibility that the rotor is just scaled down by a factor of $2.$ This was taken out by one of the edits (why do people think that's OK to do?) but the latest edit restored the "same power" condition, which I think still rules out the possibility that the rotor is just scaled down by a factor of $2$ while still spinning at the same rpm. In other words, the things Peter Kämpf took into consideration are part of the problem statement.

Using momentum theory and assumption of uniform inflow distribution, rotor thrust T can be expressed as
$$ T = C_T * ρ * A * (ωR)^2 $$
And power P as
$$ P =  C_P * ρ * A * (ωR)^3$$
From empirical data, for rotor solidity = 0.1:

Case 1: R = 13 ft = 3.96m, A = π * $R^2$ = 49.32 $m^2$, $T_1$ = 635*9.81 = 6229.4 N, assume ρ = 1.225 kg/$m^3$ and tip speed = ωR = 200 m/s
=> $C_T$ = 6229.4/(1.225*49.32*(200)$^2$) = 0.0026; $C_P$ = 0.00025; P = 120.8 kW
Case 2: R = 6.5 ft = 1.98m, A = π * $R^2$ = 12.32 $m^2$, P = 120.8 kW, assume ρ = 1.225 kg/$m^3$ and tip speed = ωR = 200 m/s
=> $C_P$ = 120,800/(1.225*12.32*200$^3$) = 0.001, $C_T$ = 0.0105,
$T_2$ = 0.0105 * 1.225 * 12.32 * 200$^2$ = 6338.6N
According to momentum theory and uniform inflow assumption, the smaller rotor delivers 2% more thrust at equal power. The rotor runs at twice the RPM and the blade is at maximum angle of attack, may even be stalling. It is the absolute maximum thrust it can deliver.
Surprised because it is counter-intuitive? So am I.
Source: Principles of Helicopter Aerodynamics by J. Gordon Leishman section 2.5


text:
Autorotation Index: The more gravity, the better?
I have been working on helicopter design and autorotation for a while. An Index used for this is the so-called "Autorotation Index" Evan A. Fradenborugh derives it in this work https://doi.org/10.4050/JAHS.29.73. Basically, it describes the ratio of available energy compared to needed energy for a flare.
In Equation 10 he presents the final result, but in Equation 9, he also provides a useable index that is non-dimensional. In this Equation, an influence is ϱ * g (disregarding surface-level density as it cancels out anyways). It seems obvious why a denser atmosphere increases the "energy" available, as you can create more lift in denser parts of the atmosphere.
However, I do not understand how g is also increasing this factor. If at all, a stronger gravitational pull should lower this factor as more energy is needed in the rotor to overcome gravity and land safely.
What are your thoughts on this?

Please avoid pseudo-answers-in-comments. If you want to answer, do so and it can be voted on appropriately. If you want to speculate what the answer might be, feel free to but you don't need to add it as a comment.

There is an analogy with gliders: long as wing AoA is optimal, there is no weight penalty to glide ratio.  The gliders potential energy, mg$\Delta$h, matches its lift requirement: mg, or pound-force.  A heavier glider glides faster, enabling it to support its greater weight.

I have found the solution.
In the Nomenclature, Fradenburgh states that $W$ has the unit of lb (thus mass). Using something that is a force as a mass is not untypical in engineering and done sometimes.
However, if we look at EQ1 and do a dimensional analysis, we can see where this leads us. $ROD$ is supposed to be descent rate, so in SI-Units $\frac{m}{s}$. Thus a dimensional analysis of the formula should lead us to $\frac{m}{s}$

The unit does not match the speed. Speed requires time which is nowhere to be seen. What if we assumed $W$ was a force rather than a mass? For this we need to multiply $\frac{m}{s^2}$ by the term below the root:

As expected, the units start to make sense.
This propagates to EQ 9, where $W$ and $DL$ appear in the denominator.
Assuming steady-state, they can be expanded to $W=mg$.
Thus, the following relation to gravity can be derived from EQ 9 (disregarding all other parameters):

Which is way more intuitive (to me).
The solution is that the Nomenclature is (at least) imprecise or (at worst) wrong, FPS as a Unit system was used implicitly, which is using pounds as a unit of force. Basically an oversight by myself.

Thanks for coming back and including such a detailed, well laid out, answer to your question.

Of course :) no problem

The author clearly employs pounds as units of force, within the coherent FPS unit system. Hence, he expresses weight and thrust in pounds, power in foot-pounds per second, and air density in slug/cubic foot...

You are right that he uses FPS. I would replace your "clearly" by "apparently", otherwise I agree. I should have seen that.

Yeah, derp, nothing about using pounds as both force and mass is ever clear to the vast majority of the world. I would have made that same mistake myself.

Just above equation 2, the direction of $g$ is defined to be up.  Therefore, this is not the normal acceleration due to gravity, but rather the opposite.

Can you elaborate on that? I am clueless about how you got to that result. I can't deduct that from the text between EQ1 an 2. Furthermore, I added an answer that (to me) seems more reasonable. Do you have any thoughts on that? If I understand you correctly these ideas would compete.

@Clex "Arrest the rate of descent" means the acceleration is up.

Reading the whole sentence you are referring to I understand that acceleration provided thru thrust "arrests" (I think I use that word correctly) is the rate of descent that would otherwise be accelerated by weight. In that case, $g$ is not defined upwards.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Do "air ferries" for civilian vehicles still exist?
In the 1950's and 1960's, there were vehicle (and passenger) air ferries operated by planes such as the Bristol Freighter. Mostly across the English Channel (England to France), by companies like Silver City Airways and Channel Air Bridge.  In other words, your car travelled with you on the same plane.
On that route, sea ferries (and hovercraft) eventually killed them off financially.
Do similar scheduled air ferries, for civilian cars and their passengers, still exist anywhere in the world?  If not, why?  Is the reason just financial, or are there no suitable planes any more, or did new safety or security rules make it impossible to operate them?
They would seem well suited to short hops like the Scottish islands, islands off British Columbia or Alaska, Indonesia, the Caribbean islands, and so on.  Especially in locations where the sea crossing is rough and unreliable.

Does this count? https://www.youtube.com/watch?v=IBu7mU1yi9g

My first flight ever, age a few weeks, was on such a service.

There are a number of suitable airplanes - C-130 Hercules, C-17 Globemaster, C-5 Galaxy, AN-124, up to the rather monstrous An-225 Mriya. Cargo versions of airliners don't usually handle vehicles - they are designed for palletized cargo while the above cargo planes have ground-level ramps.
If you want to fly with your car somewhere it's just a matter of calling someone who owns one of these airplanes and putting down your no-limit credit card.
As you surmised, the reasons they don't operate scheduled service is financial - boats are much cheaper, and not enough people want to go where a boat is impractical.

or people who would use such a service would just get a second car at the destination, much cheaper in the long run

The people that participate in Gumball 3000 certainly have the money (and the cars) to pull such a stunt off: http://uniquelytoronto.wordpress.com/2014/01/09/gumball-3000-rally-2014-dates-and-locations/

It's also worth pointing out that the specific route above (England<-->France over the English Channel) has been largely superseded by the Channel Tunnel (and EuroTunnel trains that will happily take you and your car back and forth in less than an hour). That's tough for an air carrier to compete with on both time and cost...

To answer the question directly:  No, there are no scheduled automobile ferrys at this time.  Can certain aircraft carry something the size and weight of a car?  Of course, see other replies here for a list of some of the planes with this capability.  Why not?  The answer is, as you suspect, financial.  In most circumstances it is cheaper to buy or rent a vehicle where you need one.


text:
Is the freestream dynamic pressure higher during cruise or during landing/takeoff?
I was trying to calculate the forces generated by an uncommanded flap deployment during cruising flight ($M_\infty=0.85, Alt.=36kft$) in comparison to the forces say, during landing. Assuming similar geometries, the forces will be proportional to the freestream dynamic pressure. If $q_\infty=\frac{\gamma}{2}p_\infty {M_\infty}^2$, the Mach number will be higher at cruise but the static pressure will be lower due to altitude. Thus, it is not immediately obvious to me which flight condition has the higher dynamic pressure. Does anyone have reasonable numbers for $q_\infty [lbf/ft^2]$ at landing vs cruise for a modern commercial airliner so that I can compare?

$lbf/ft^2$? http://xkcd.com/526/

I don't understand the problem. I want pounds-force per foot squared.

The equivalent speed is about twice as high (the calculator linked by fooot gives 266 knots for your case while 130 is reasonable landing speed) and since equivalent speed is proportional to square root of dynamic pressure, it means about four times higher dynamic pressure at cruise.

Here is what I am getting:
Static pressure:
Sea level: 101300 Pa
30,000 feet: 30100 Pa
Mach:
Sea level1 (150 kt): 0.22
Cruise2: 0.85
Ratio of specific heats: 1.4
Using the equation:
Landing: $3432.044 Pa = 73.68 lbf/ft^2$
Cruise: $15223.075 Pa = 317.94 lbf/ft^2$
So the high Mach in cruise more than compensates for the lower static pressure. This calculator agrees.
1Landing speeds for large jets are generally between 120-150 knots (Mach 0.18-0.22). Smaller planes around 60 knots (Mach 0.091). You can convert to Mach by hand or using a calculator like here.
2Cruise speeds are generally between 240-280 kias (knots indicated airspeed) for large jets, which translates to around 450-500 knots ground speed and Mach 0.75-0.85 (depending on altitude and conditions).

Well for planes that do Mach 0.85 at 30,000 feet it's generally around landing speed...

This is why I was asking the question, I really don't know what a reasonable landing Mach is to two significant figures...

@fooot: Cruise speed is not around landing speed, it's about twice as high. The calculator gives for the listed conditions equivalent airspeed (the one most closely related to dynamic pressure) 266 knots. Which is about exactly double (130 knots is reasonable landing speed) and that means four times as high dynamic pressure.

The aircraft needs flaps in order to be slow enough for take-off and landing. This means the lift coefficient $c_l$ is larger in the landing configuration, and the dynamic pressure is proportionally lower, because lift is the product of $c_l$, the wing area and the dynamic pressure. Since the mass of the aircraft is roughly the same at the end of cruise and during landing, this product must be the same in both conditions.

Real-world examples: Double slotted flaps plus slats will give you a $c_{l\,max}$ of maybe 3.2, so your flight $c_l$ is around 2 to have enough safety margin. Cruise $c_l$ for the same airplane would be between 0.3 and 0.4.

Now with your more precise question I can give a more precise answer. Let's say wing loading is 500 $kg/m^2$, then your dynamic pressure in the landing phase will be 2600 $N/m^2$ versus 14000 $N/m^2$ in cruise. I guess this has not been obvious because you were stuck with imperial units. Do it with metric units, and things should be immediately clear. Formula: $$q = \frac{m \cdot g}{S \cdot c_l}$$

I already know all of this. Essentially what I am asking is whether or not the flaps would be ripped off the airplane if they were deployed during cruise. It is not intuitively obvious for the reasons I outlined in the original post...

@BrysonS. That's really more a structural question than an aerodynamic one (and the answer is different for each aircraft - one may be able to survive deploying flaps at cruise airspeeds, while another design may have the control surface torn from the aircraft).

Normally, the maximum flap operation speed is 1.7 x stall speed. Speed here means calibrated speed, and your calibrated speed in cruise is higher by a factor of 5. So yes, the flaps will be torn off, no doubt, or they were designed much too heavy.

@Peter Kampf. Yes I think this is generally what will happen, but I was trying to put a number to it. I will use the 1976 ISA to get a reasonable static pressure for both landing and cruise, but I still need a reasonable number for landing Mach (with full flaps deployed). Thoughts?

@Bryson S. Don't make it more complicated than necessary. Mach in the landing configuration is irrelevant, because it is so small (warmer air plus slower speed). If you need a value, use 0.15. Static pressure is also not needed, only density, and that is part of the dynamic pressure already.

Your answer lists the wrong units for pressure ($N/m^3$ instead of $N/m^2$), but I do like the idea of taking Mach number out of the equation and talking in terms of wing loading and $C_l$.

@Bryson S. Thanks, corrected.


text:
Why do autogyros exist?
This question has been intriguing me for some time. In Autogyro the rotary blades merely acts as what fixed wings act in normal aircrafts to generate lift. So why not simply have a fixed wing instead of rotary "wings"? What advantage do this rotary nature of wings provide compared to fixed wings? In both ordinary planes and autogyros the propulsion is generated by the front or aft engine.

@ManuH the article contrasts autogyro but the main feature that was mentioned by Robin was not there which is what I was looking for.

you should explicit this in the description of your question This show your current level of research (that help us adapt the details of the answers) and avoid repeating facts that are already mentioned in other questions.

I am not sure what you mean I did explicitly mention what is the purpose of rotary wings over fixed wings.

I mean, you should explicit what you didn't find in the (almost) dupe.

Autogyro blades have a higher airspeed than the aircraft that is carrying them, so they can generate lift even when the aircraft speed is low. They fly as if they had a wing the size of the whole disk, not just the blade area - yet can be stored or transported easily.
They can land (but not take off) almost vertically, which is a useful safety feature in the event of an engine failure.
Having said that, they aren't common because they don't do anything useful significantly better than the other options, except perhaps being 'different and interesting'. Flying slowly isn't terribly useful.

Interesting fact. Gyrocopters were being considered by the Bureau of Land Management as a cheaper alternative to fixed-wing aircraft and other rotary-wing aircraft. They cost less to operate than a helicopter and have a lower procurement cost than most brand new airplanes. I don’t know how far they got in the project development phase. I think they got the idea from airborne cattle herders in Australia.

"Flying slowly isn't terribly useful" Actually it is useful. Autogryo's are often used for wildlife surveys specifically because they fly so slow. They are also used for cable surveys (they are a lot cheaper to operate than helicopters) and even for anti-poaching operations in Africa.

+1 for mentioning They fly as if they had a wing the size of the whole disk, not just the blade area Now it makes lots of sense.

Actually, there were and are gyroplanes that are capable of "jump" takeoff -- they have collective pitch control and a means of spinning up the rotor, then use the momentum of the rotor and a fast collective increase to "jump" off the ground; with the engine already powered up, then then transition into forward almost instantly.  There was a commercial passenger model in the mid-20th century that could jump 6-10 meters when loaded.

Jump takeoff gyros are today a rarity. I have been flying gyros for twenty years and have never seen an active jump-takeoff gyro. They can, however, be seen in museums...

The autogyro was invented as an aircraft that cannot stall - a friend of the inventor was killed in an aircraft stall accident.
The usefulness of the autogyro is that it was the necessary step on the way to inventing the helicopter.


text:
Do aerodynamic forces depend on the materials of the wing surfaces?
I think that statement is true, but in the book "Fundamental of aerodynamic", page 35: the aerodynamic forces R don't depend on the skin material, 5 variables are mentioned but none of them is relate to skin material of wings. So why is that???

Not on the material, but on the surface roughness and flexibility.

yeah, I mean that, surface roughness but the question is still the same. why are 5 variables not related to the surface roughness ?

Effect of Surface Roughness on Characteristics of Aerofoils N.A.C.A. 0012 and R.A.F. 34, written 81 years ago.

don't simply post an image full of text, tape the sentences that are needed to understand the question and eventually give a link to the book.

Generally, skin materials will impact the local deformation of the surface of the airfoil (think parachute).  They also can impact the local boundary layer, with things like surface texture (think Reynolds number changes). Generally speaking these effects are tertiary, and not a prime mover in R.
If you are looking for something more comprehensive, then I will have to break open the book.
Addendum #1
Understood the question is redefined. In this case, I agree with @mins and his commented reference to the Jones and Williams paper. You can find it readily on the net.  It addresses the surface texture and corresponding Reynolds number impact.  There are some recent works which take that further and model it nicely, but that is beyond the scope of the question.
In general, there are two regions of Reynolds, laminar and turbulent, and how they impact drag is different.  Perhaps the question should be reformulated.

sorry, I mean surface roughness, forget about the deformation things. I think drag depend much on surface roughness but the page did not mention that. that confuses me

There is induced drag and parasitic drag. Parasitic drag is impacted more by surface roughness.

Surface roughness has a minor impact, except in a range of Reynolds numbers mostly encountered in General Aviation. Only when you use an airfoil that has been designed for laminar flow but experiences an early transition will the performance be impacted significantly.
The book you quote assumes a certain minimum standard of build quality: Flush riveting, smooth surface and a solid, stiff structure. Once you move to more advanced topics, more variables will need to be added, such as elastic deformation in flight or substandard surface quality. At the introduction level some simplifications are justified. The lessons derived from this reasoning are valid and valuable. When you progress, they will be expanded with more complex concepts. Trust me.


text:
What is this strange structure, of a type often seen by pilots in hilly regions?
I spotted a strange structure near a few paragliding spots in italian Alps. It's a screen about 10 meters wide, 6 meters tall, mounted on concrete base on ridge top. Someone said it's a training target for fighter jets, is it true? If so, is it targeted with real ammo or some kind of video-based training system?
Pictured: Monte Dolada, 46°11'34.68"N, 12°20'25.41"E

It's a reflector for microwave links, a passive repeater.
More info here (article in Spanish): http://notascordobesas.blogspot.com/2015/01/el-repetidor-pasivo-de-microondas.html


text:
Could atmospheric effects cause a PA-30 to break up?
About a week ago, a Piper PA-30 crashed into a residential area just a few miles from where I live. An engine was found to have landed a 0.24 nm from the impact site. Sadly, the pilot of the aircraft was killed. Thankfully, however, in spite of the crash occurring in a crowded subdivision, nobody on the ground was injured.
Now, to the heart of my query:
The NTSB preliminary report indicated an in-flight breakup of the plane. There was not an active thunderstorm in the area, but with 30 minutes or so a small cell had developed (source: me, I do live nearby). Are smaller aircraft subject to atmospheric effects that could cause an in-flight breakup? If so, what sort of atmospheric conditions, assuming a skilled pilot who wouldn't purposefully fly into known bad conditions, would be necessary to cause such catastrophic damage to a small craft like a PA-30?

This it too broad of a question for a simple answer.  It would require a detailed Failure Mode Effect Analysis and that can be a pretty big document

I think your answer is to wait for the accident report to come out. (I'm not in the US so don't know what sort of reports the NTSB put out or where they publish them.) If this is about what sort of accidents can happen to light aircraft leading to loss of control, it might be too broad for a single answer.

@Andy Right this way for NTSB accident reports...

Was the 1971 crash necessarily the same plane? The [NTSB summary of that incident]((http://ntsb.gov/_layouts/ntsb.aviation/brief.aspx?ev_id=61474&key=0) doesn't list the plane's serial number and my understanding is that tail numbers aren't unique. Isn't it possible that the owner of the plane that crashed in 1971 replaced it with another of the same model and gave it the same tail number.

@DavidRicherby That... Is an excellent question. Do you have a reference that tail numbers are re-used, and would be allowed to be re-used on an aircraft of the same model?

aviationdb.net shows the aircraft involved in the 1971 accident also had serial number 30-470. It was described as "destroyed" ...."Occurrences: STALL; Phase of Flight: LANDING - OTHER; Causes: IMPROPER COMPENSATION FOR WIND CONDITIONS, FAILED TO OBTAIN/MAINTAIN FLYING SPEED,RAIN; Narrative: LDG RWY 31,DRIFTING R OF RWY,TIGHTENED APCH.". The pilot apparently survived, so it was a good landing.

This question is to broad to be answered here but there are some good questions with in it. If you are curious about smaller plane glide ratios and maneuverability or how engines are mounted and subsequently can shear off those topics may make for good individual questions.

@RedGrittyBrick, that is in fact the same serial number as the plane that crashed here a week ago. Maybe I should ask a new question about that - what takes a plane from "destroyed" back to air-worthy?

@Dustin, you may find an answer to that question here

@DeltaLima, thanks. I had mistakenly assumed what the "destroyed" nomenclature meant; I see now it's more like when we say a car was "totaled" - it's a "total loss" of value vs. repair cost, rather than a total loss of the vehicle itself.

Are you saying the thunderstorm developed within 30 minutes after the accident?

@TomMcW yes, a small, sort-lived cell

Hmm... I'll be curious to see if they conclude weather was a factor.

Atmospheric conditions do not cause aircraft to break up in flight.
Improper aircraft handling in response to bad weather can cause an aircraft to be overstressed and then breakup.
So can improper handling due to disorientation.

Is that likely to be the case with a pilot with "a Federal Aviation Administration (FAA) commercial pilot certificate with airplane multiengine land and instrument airplane ratings [...and...] pilot privileges in single engine land airplanes"? I don't know if over 900 hours' flight time and his licensure is enough to rule out improper handling, I'm asking honestly.

@Dustin The more experienced a pilot is the less likely they would be to mis-handle the aircraft, but there are many cases of extremely experienced pilots doing so. It can happen to anybody if they get complacent or confused

@TomMcW thanks. This pilot seemed to be one that was "connected" with their aircraft - took it to shows and such - but of course there's no way for me to know what level of experience that suggests. I have wanted to pursue my private license for years, and this accident has made me sit up and take note of new variables I hadn't considered before. The responses on this thread have helped me to understand more, and I appreciate everyone's willingness to share their commentary.

Given that this is a direct "answer" to my query, I'm going to mark it as such. Thanks, everyone!


text:
How (and when) does Heathrow alternate arrival paths?
I've noticed that flights over Richmond (particularly arrivals) seem to have significantly reduced over the last few weeks. This is based on looking at Flightradar, as well as listening to the noise levels overhead.
I believe that Heathrow ATC alternates runways, holding patterns, and directions, but I'm finding it difficult to understand how and when these changes happen.
When do these flight path pattern changes happen, and potentially when will flights overhead Richmond might increase again?

Welcome to aviation.SE! This question may be relevant.

Heathrow has an extraordinarily elaborate runway alternation programme, designed to spread out noise as evenly between its different neighbours as possible.

Daytime runway alternation follows a daily cycle and a fortnightly cycle. Alternation starts at 6am and continues till the last aircraft departs at the end of the day.

In the morning we use one runway for take-offs and the other for landings. At 3pm – roughly halfway through Heathrow's working day – we swap over
We continue this morning and evening pattern of runway use for the rest of the week
The following week we switch completely. What we did in the evening during the previous week, we now do in the morning. And vice versa

Regarding planes over Richmond (which is situated under the approach path to runway 27L):
During the week starting 8th of April, the programme states that 27L was used for landings from 6:00 until 15:00, resulting in a lot of traffic over Richmond during the morning, but little to none in the evening.
During the week starting 15th of April, the programme states that 27L is used for landings from 15:00 until the end of the day.
During the week starting 22nd of April, they will switch over again to land on 27L from 6:00 to 15:00.
You can read much more about the programme here.

ATC will changes runways depending on the direction of the wind. Heathrow has 2 strips:

27 Left/09 Right
27 Right/09 Left

It is a good thing to land into wind, so if the wind is coming from the east the 09 runways will be used. In London the prevailing wind is more often coming from the southwest, which is why the 27 runways are more often in use. However, historically you'll find that the winds in March and April show a lot more variability, with northeasterly winds much more often, which has been true this year. A wind from the east will mean the 09 runways will be used. Departing airplanes tend to get turned and dispersed relatively quickly, so you'll fewer airplanes taking off over Richmond than on approach.
Richmond is directly under the approach path for 27 Left. Heathrow will often switch the left and right runways in order to reduce the noise impact on local residents. The hold points haven't changed that I know of, but there are several of them and they use different ones depending on the runways in use.
So the combination of wind directions this time of year and the switching of runways means you'll see less airplanes over Richmond this time of year. But never fear, by July the wind will be back to the southwest most of the time.


text:
Where do LAX commercial cargo planes land/take off?
At the Los Angeles airport, do commercial freight planes such as a 777F, 737F etc. take off near the Lufthansa / FedEx / DHL cargo areas from the 25L /25R runways towards the ocean ?
Do they land in the same direction but farther down the runway?

They takeoff and land into the wind, just like any other plane.  They can takeoff and land with a tailwind, but it can't be a very strong wind.

I haven't operated out of LAX for some time, however, from the early 80's to the mid 90's LAX operated with some generally predictable patterns. Freight carriers were handled in the same way as all other carriers. Takeoffs were USUALLY conducted from 25R and 24L (Inner Pair). Landings USUALLY conducted form 25L and 24R (Outer Pair). During weather with significant wind from the West TO's 6R & 7L / Landings 6L & 7R. Late at night during normal op with minimal wind from the West the 6's and 7's would be used for landings to mitigate noise over the city.

Large airports like LAX see hundreds of planes departing or arriving every day. They manage traffic flow in the entire area (class Bravo airspace) to optimize their traffic. Cargo aircraft will be treated just like any other aircraft. Which runways are in use for departures and which ones for arrivals, depends on the current winds and preferred configurations of the airport. Due to the high traffic, the airport will not deviate from this configuration to accommodate a single cargo aircraft, which would then be closer to its cargo area. Cargo aircraft will then taxi to their respective cargo areas just like passenger aircraft taxi to their gate on the ground.


text:
What is the purpose of mixture control?
What's the purpose of mixture control? Is it to control the amount of fuel provided to the engine with which it is associated? If that's the case, what's the purpose of throttle lever then?

Does this answer your question? https://aviation.stackexchange.com/questions/233/how-does-the-mixture-of-an-engine-affect-the-engines-operation

Yep, clearly a duplicate...

@MichaelHall So, if you thought so why OOI did you not vote to close?

The throttle directly controls the air admitted to the engine.  In a carburetor (and most gasoline engines with fuel injection as well) this is done with a "butterfly valve" in the intake that rotates from a full throttle position parallel to the airflow, to an idle position that nearly blocks the intake (permitting only a small fraction of the full throttle flow).
Mixture control then adjusts how much fuel is added to that airflow, as a ratio (that is, the mixture compensates for throttle condition, either via manifold pressure as in a carburetor or via mass flow sensor in at least some fuel injection systems).  This is important because most gasoline aircraft engines are air cooled -- which means they're partially "charge cooled", with excess heat carried away to some extent by unburned fuel in the exhaust, so adjusting mixture not only adjusts power produced at a given manifold pressure (set by the throttle), but also controls heating (richer mixture will usually give cooler exhaust gas temperature, EGT, which is a proxy for combustion temperature) -- therefore you'll set mixture rich (more fuel for a given airflow) for takeoff and climb, and then "lean out" by manifold pressure and rpm (and/or EGT) for cruise economy.
In older carburetor systems, this was done with a needle valve in the main carburetor jet, which adjusted the flow area where the gasoline was metered into the airstream (you can see this in detail by examining a model airplane engine's carburetor, where the needle valve is adjusted by a thumbscrew, only on the ground).  With fuel injection systems, there's either an analog computer that meters the amount of fuel according to the airflow, or a digital computer controlling the injector(s) that does the same job with more ability to compensate for things like sudden throttle changes, temperature, outside humidity, and even fuel quality.
Almost all modern automobile engines have the latter, which is combined with the computer that manages many other functions in the car; aircraft engines are usually more resistant innovation (due to the huge cost of certification), so injections systems there are often 1970s technology, if the airplane they're in is new enough even to have that.


text:
Why are planes flying in a zigzag path over Long Island when approaching JFK?
Every time I've flown into JFK Airport from the east I've noticed that the plane takes a weird zigzag path over Long Island. Why is this? Is it because of noise laws?

Just a guess, but it's likely a combination of routes following existing navaids and fixes (which don't line up in straight lines) and keeping inbound JFK traffic away from Long Island Mac Arthur (ISP) traffic.

@Gerry: Why does JFK traffic have to make way for ISP traffic, rather than the other way around?

@Sean  Because it's called the National Airspace SYSTEM.  It's about optimizing everyone's use of the airspace.  The airspace right around an airport is the most congested so it makes sense to keep aircraft not going to or from that airport out of the way.  It goes both ways.  If you're in northern PA headed for KISP, you're not going to get a straight flight through the NYC airspace.  You have to avoid interfering with LGA, JFK, EWR, and TEB traffic.

There are 3 STARs (Standard Terminal Arrival Routes) into JFK that cross over Long Island:

PARCH 3 RNAV ARR:

PAWLING 2 ARR:

ROBER 2 ARR:

(Jeppesen charts)

None of these have particularly pronounced zigzag patterns, so what you describe is probably not coming from the waypoints of the arrival procedure.
Looking at flightradar24.com, we can currently see BAW1 approaching JFK:

This already looks a bit more like a zigzag. It would seem they are arriving on the PARCH3 or ROBER2 until ROBER and then received radar vectors onto final approach for runway 31L/R. These vectors are typically given by ATC (Air Traffic Control) for spacing reasons.
When going back to yesterday evening using flightradar24's Playback feature, one finds much more pronounced turns over Long Island (thanks to ymb1 for suggesting this):

Here, aircraft are approaching runways 22L/R. You can see that there are also aircraft from the South and West joining the queue to land and ATC needs to make sure there is enough separation between the aircraft. The easiest way to achieve this is by instructing them to fly in a certain direction ("radar vectors") until there is enough space to fit them into the queue, which results in such a zigzag pattern.
Such turns can also be more localized than shown here. See e.g. Why did my LGA-ORD flight make an S-shaped turn round the time it was passing a storm?, where a flight into Chicago made an S-shaped turn during the descent, which would also feel like a zigzag. Such instructions can also be given to aircraft while they are still flying on the STAR over Long Island, especially when there is a lot of traffic during peak hours.

If I'm flying into JFK it's usually around 10 pm, if that matters. The entire approach is over land and it goes from top to bottom to top etc.

@MattSamuel That makes sense. You can see that all arrivals from the East fly over Calverton (North side) and then ROBER (South side). If runways 22L/R are in use, they have to fly back to the North side afterwards to join the final approach. I found a nice example of this from yesterday evening (I think around 11 PM, but looks exactly like what you describe).

The time might have been 11pm-12am. I just remember that it was a god awful hour with two very young children who usually go to sleep at 7 pm.

Why do all of these arrivals fly over Long Island at all prior to final approach, rather than coming in over Long Island Sound or the Atlantic?

@Sean The North Atlantic Tracks are mostly coming from the North East, so it is simply the shortest route for a majority of the arrivals.

@Bianfable: ...yes, so why are they arriving over eastern/central Long Island rather than over Long Island Sound?

@Sean I don't know. I can tell you by looking at the high enroute airways in the area that there are no airways from anywhere South of Boston leading into the Long Island Sound. They all end up at the South East corner of Long Island, but I'm not sure why...

@ymb1: Except that you don't have to overfly a VOR (or any other type of navaid) in order to use it; waypoints can be defined just as easily by intersections of radials, or by DME distances along a radial, as they can by the locations of the navaids themselves.  Just because a VOR is on land doesn't mean that it can't serve overwater flights.

It's worth pointing out that the entire New York/New Jersey/Philadelphia airspace design is a giant puzzle that has to fit together between the region's airports, accommodating flights to/from various directions under different weather configurations. Flights use certain bits of airspace because other flights are somewhere else. Looking at FlightRadar24 right now, I see flights over Long Island Sound for White Planes and New Haven. This allows flights into those airports to fly over the JFK arrival stream and then descend over the sound.

When I'm flying into JFK, it's from Guyana, which is in South America, and they still fly over Long Island.


text:
Why do some airports have more than 2 runways, when LHR can get by with 2?
I was looking at aerial imagery of various airports and noticed how some have a large number of runways (4-6), while others have a small number.
Specifically, I noticed that Heathrow (LHR) only has 2 runways, but it is the busiest airport in Europe by passengers. In comparison, the following busiest airport have many more runways (Paris Charles-de-Gaulle (CDG) has 4, Amsterdam Schiphol (AMS) has 6), but fewer passengers.
Why do some airports have more than 2 runways, when LHR can get by with 2?

but fewer passengers More passengers != more planes. What if Amsterdam has more cargo traffic than Heathrow? Don't know, but you might want to check the actual plane number serviced.

Amsterdam does have 6 runways, but they only ever use 2 or 3 at once for commercial traffic, since you can't just independently use runways that intersect, for example.

Heathrow has several more runways, it's called Gatwick, Luton, Stansted, and London City airports...

Please do not substantially edit a question to invalidate existing answers.  If you need to make such a change, just ask a new question.  Thanks.

It is a question related to politics more than aviation

Heathrow barely gets by with two runways; flights into Heathrow routinely have to hold for hours before they get a chance to land, and they're building a third runway precisely because the two they have now are barely sufficient.

Heathrow is somewhat of a standout when it comes to volume. Your wording, that LHR "gets by" with 2 runways is apt. By movements, it is the busiest 2-runway airport in the world. By passenger volume it is second only to Dubai Intl. (Note, though, that DXB is home to Emirates, who has the world's largest fleet of A380's. They are the only airport that averages over 200 passengers per movement!)
Aircraft movement is the best way to look at it from a runway usage standpoint. The following chart is sorted by aircraft movement.*

As you can see, the top 3 all have 5 or more runways. There are 7 large airports with 5 or more rwys and they all fall in the top 16. The top 16 is rounded out by 5 airports with 4 runways, three with 3 runways and... LHR. The next 2-rwy airport on the list is Benito Juarez Intl. in Mexico City clear down at 21st on the list.
Any busy airport with 2 runways has to make those runways work as tightly as possible with little room for error. Otherwise they end up with delays. If I sort the list by movements per rwy, all of those with 2 rwys rise to the top. 10 of the top 15 have 2 rwys.

So, how does LHR pull it off? How do they manage to pack all those flights in to just two runways?There's a downside to handling that kind of volume with so few runways. The following collage gives a clue as to what that is.

Those are just a random pick of arrivals at a couple of random times yesterday morning at LHR. There was no significant weather at the times. As you can see, you just don't arrive at LHR without being put in a holding stack. Every single flight I could find had to do at least one racetrack. At any point during the day, in places like Chesham, Effingham and Brentwood, if you look up there will be as many as 6 or 7 aircraft doing racetracks waiting to be cleared into Heathrow. That makes for very efficient runway usage when there are 20 planes waiting in line, but it's terribly inefficient for the airlines and passengers to have to spend so much time burning fuel circling around London.
On the flip side, building more runways has a down side too. It will get the planes out of the air quicker, but, as happens at Schipol - which has 6 runways - you can end up with a long taxi back to the terminal.

All data was taken from Wikipedia.

'nearly an hour' seems to be an exaggeration, here they mention 20 minutes, and wiki says 15 minutes.

The long delay isn't caused by the 6 runways, it's caused by the orientation of that 6th runway. If it had been rotated by 20 degrees CCW, the south end would have been a lot closer to the terminal, but the north end would have been near the local golf club.

@ROIMaison I've never been there. Just going by what I've heard others say, and they were probably exaggerating. Looks like probably 15-20 min. Edited to remove the hyperbole

@MSalters Looking at the satellite it looks like the 18R flight path runs pretty much over the two nearest golf courses I can see. I see the path over sparsely populated area to the north. But 20° CCW would put the path right over Haarlem. So it's certainly aligned for noise abatement reasons.

@ymb1 That's a very cool video! That's what it would look like if airplanes were spiders :)

@ymb1 wow, that flight from Stanstead that took off east, turned southwest, then south toward Gatwick then turned back...

Don't assume that a higher number of individual runways means a higher possible throughput. The runways at LHR can be used at the same time - this instantly puts them at an advantage. Many older airfields have a typical A shape runway, with 3 intersecting runways - only one runway can ever be used at once in this configuration. Other factors would include noise abatement and licensing restrictions - airports are often controversial, with flights artificially limited by legislation.
Beyond that, it's possible that not all runways are capable of operating with heavy (physically) traffic, may not have the same ILS capabilities and so on.
Finally, just because an airport COULD host more aircraft than LHR doesn't mean it will. LHR is an exceptionally busy hub airport operating at almost maximum capacity all year round. I could build a 20 runway airport in Iceland and wouldn't come close to Heathrows throughput.

Even two parallel runways can’t necessarily be used independently; e.g., the two main runways at FRA are too close for really independent use, which necessitated the addition of two other runways.

These are relevant points, but I don't see this as answering the question. Runways cost money, so why build them if they are not necessary?

@Thomas: So that one can be closed if required (accident, maintenance, conditions) and the airport left open?

@Thomas The question was changed since I answered

@Dan Thanks for the great answer. I did change the question according to a comment, but your answer does bring up important points that I do think contribute to the answer of the new question. Thanks!

To complement Dan and TomMc:

Having perpendicular runways like in JFK allow for less cross wind take off / landing
Having multiple runways with less traffic allow more flexibility if one runway was not usable (tire burst at Gatwick few days ago caused a lot of diverts / delays / cancellation because only one runway was open)
Some airports expanded and created new longer / larger runways for bigger planes


text:
Where can I get flight departure change data?
On sites like flightstats.com, there is an "event timeline" view that shows changes made to the scheduled departure or equipment for a commercial flight. I assume this information is being pulled from some central system / API.
Here's an example, from today
Jul 26  05:00   1:00 AM 6:00 AM Gate Adjustment
TAIL Changed From N14106 To N14102

Jul 26  14:27   10:27 AM    3:27 PM Gate Adjustment
Departure Gate Changed From 125 To 128

Jul 26  20:23   4:23 PM 9:23 PM Gate Adjustment
TAIL Changed From N14102 To N34131

Jul 26  21:30   5:30 PM 10:30 PM    Time Adjustment
Estimated Gate Arrival Changed From 07/27/16 07:30 AM To 07/27/16 07:31 AM

Jul 26  21:31   5:31 PM 10:31 PM    Time Adjustment
Estimated Runway Departure Changed To 07/26/16 07:50 PM
Estimated Runway Arrival Changed To 07/27/16 06:54 AM

Is there a programmatic source for this data? Perhaps with additional metadata not exposed on flightstats.com? I understand they do have an API themselves, which does not seem suited to hobby uses, so I'm looking for another source. Are these departure changes submitted to FAA/ICAO? Are they airport-local only?

Welcome to the site! You might want to review this question and this one to see if they're relevant.

Thanks, @Pondlife - I did check those, and while I have an additional interest in ADS-B - I have a Raspberry Pi dedicated to just that! - I am wondering if there is a source for specific data regarding flight delays before departure. AC changes, gate changes, departure delays, etc. Thanks for checking in so quickly :)

@mins Fair enough; is there a preferred procedure for reposting there?

I'm voting to close this question as off-topic because because it belongs to travel.stackechange.com

Welcome Dustin, you can flag your post (flag button) explaining the request to move to Travel.SE, moderators will look at it. Enjoy. EDIT: In addition you may search on SO.

@mins Thanks for the tip. I've done just that.

@mins I agree that it might be borderline for us, but I don't think that this question will have a better fate at travel.se

@Federico: Yes, I edited my comment to mention SO too, SO seems to be the best choice. I didn't look at the API in depth, but the answer may be there, and someone on Aviation may want to dig into that too.


text:
Why are commercial propeller aircraft smaller than commercial jet aircraft?
I've read that turbo-prop aircraft are more efficient than jet-aircraft. Browsing this Wikipedia page it looks like in general, jet airliners burn less fuel per passenger. I imagine this is due to jet aircraft typically carrying more people, and flying longer in cruise. Is there some fundamental reason why traditional propeller designs couldn't be scaled up to 150+ passengers and trans-oceanic ranges? If it was possible, would air travel become substantially more efficient per passenger?
Edit: Perhaps a better phrased question would be: are turboprop aircraft always more efficient, regardless of airframe size? Or is there some regime where modern turbofans take over? I'm specifically asking about fuel efficiency (i.e. ignoring noise / speed).

With great thrust comes great MTOW.

It all boils down to cost. Can an airline make enough money from passengers using their turboprop 8 hour routes when a jet powered aircraft can cover the same route in 4.5 hours?

why traditional propeller designs couldn't be scaled up to 150+ passengers and trans-oceanic ranges?

Well, they can. One of the more fundamental problems is extracting enough power from turboprops, that is, building a large turboprop engine. Turboprops are necessarily geared, and designing a gearbox to pass such an enormous power and torque is a big challenge. In the Tu-114/Tu-95 engines (which were the most powerful turboprops ever built), the gearbox weighed more than the rest of the engine. After some point, such gearbox will eat up all the benefits.
The same applies to the propeller: it is quite difficult to convert high power to thrust while maintaining a reasonable size.
The new generation geared turbofans potentially offer some of the advantages of turboprops: they need less gearing, yet their fan has much higher thrust density.

Cool, thanks for the info!

There are a number of problems:

Speed

Traditional Propeller aircraft are much slower than jets. Propellers decline in efficiency as they increase in speed.

Noise

For the same thrust, propellers are noisier then jets.

Development Cost

There have been efforts for propellers but so far they have not been justified by economics.
GE36 Engine


text:
Could the vertical stabilizer be removed on the NASA X-57?
I was reading this article about how NASA will develop this new X-57 X-plane which has many tiny electric motors mounted on the wing.

Now, I wonder: How come this aircraft still has a vertical stabilizer? I thought that one of the points of having many tiny electric motors was that you needn’t worry about an engine failure (one engine fails, you shut down the one on the opposite side; you still have 12 working, no big deal). You probably don’t have to worry about lateral stability either, because those motors can probably easily stabilize sideslip oscillations (a bit like on a quadrotor, where you can maintain a stable hovering position thanks to accelerometers and fast control feedback).
Am I missing something?? If so, isn't there a way to downsize the stabilizer a little bit? How would you go about calculating that?
EDIT: I realize the problem of power failure, but couldn't it be just avoided by carrying lots of independant battery packs (instead of a central big one), so you can be sure that you statistically never lose control of all your engines?

Well, during a full power failure (complete battery failure) the aircraft would be a glider, which could benefit from a vertical stabilizer.

Very few finless aircraft have been built using engine thrust (and maybe spoilers) to control yaw; they were all military too I think. A tail fin really is a simple, reliable way to control most "normal" aircraft, including this one. It will work even without engines, as Steve H just said.

I agree, the fin is a great solution; but a heavy and draggy one too... I just thought NASA could be a little more ambitious in their design :-)

If you want ambitious look at the B-2 Spirit. But if the controls go down, you have to eject - which has happened.

So lots of spare batteries, redundant electronics and control systems is better than just having a rudder?

The batteries don't have to be spares, just isolated (like different fuel tanks). Electronics should be redundant anyway for safety. The fact that you can remove the rudder is a consequence of a smart choice of propulsion architecture...

Actually, during a battery failure you might want to turn two of the engines into Ram Air Turbines. That's fairly easy with electrical engines.

Differential thrust is not a good way to control the yaw of an airplane. For one thing thrust control has to be extremely quick, if it takes even a second between a control input and the thrust response then you won't have adequate control over the airplane. For another you are entirely dependent on having engine power available for stability, if you lose your electrical system or your battery goes dead then so are you!
Another reason differential thrust is not a good idea is that at full power the only way to create differential thrust is to reduce power on one side. If I'm on climbout and I need to yaw I will lose climbing power, this means I would have to reduce my flight path angle to maintain airspeed. If I'm close to stall speed a sudden reduction in power could make me enough airspeed to stall. Not good. The only way to really mitigate against this is to increase motor power and restrict the maximum power a pilot can get. This is extra weight and cost.
Lastly, the X-57 has the engines all down the wings, this means that when the engines are going they will be producing airflow over the wing. Differential thrust means that the engines are forcing more air over one wing than the other, the wing getting more airflow will create more lift, this has the following undesirable characteristics:

A yaw input will have to be coordinated with reverse aileron to maintain level flight, a strong yaw input would need a stronger aileron change. This means badly coordinated controls and would not be very intuitive
At high angles of attack, for instance close to stall speed, sudden reduction of thrust over one wing could stall it while the other wing keeps flying, this would lead to a snap-roll and possible loss of control. In short it's dangerous.

Comparing quadcopter technologies to the technology going onto the X-57 is a bit of a misnomer:

Quadcopter motors are small and the blades are small compared to what will be used on the X-57, the ones on an airplane will be much bigger and will have a much larger response time as a result due to the longer moment arm
A quad has a higher thrust to weight ratio than the X-57 will, so less power available for differential thrust

Don't you think that tiny electric motors could be indeed very reactive (probably way faster than a second)? Gyrometers give you the yaw rate instantly as well... (like on a quadcopter)

These motors and propellers are quite a bit bigger than those on your average consumer grade quad copter, and even those on a large one used for TV/Movie filming, therefore, they have quite a bit more inertia than those tiny motors do. Even if the electrical impulses & computations take milliseconds, changing the actual rotational speed of the propeller(s) could take a second or longer, simply due to physics. As we all know, physics always wins.

@HughKeller also bear in mind the yaw response of quadcopters is not controlled by the thrust of the rotors - its controlled by changing the angular momentum of the rotors. It's like having a reaction wheel. Instant response (limited only by motor+rotor acceleration).

Another very good reason for not changing the tail surface at all is that this is a research aircraft--it is not an ideally optimized aircraft.  By basing it on the Tecnam fuselage with no changes, NASA can get a baseline number for the drag reduction allowed by this small high aspect ratio wing vs. the original, much larger wing on the Tecnam twin engine aircraft. They are purposely isolating this single variable.  Indeed, a paper authored by someone on this project gives an airframe drag reduction estimate of 1.5 times compared to the parent aircraft; they believe an optimized design could reduce drag further.